Respuesta :
Answer:
The area of the sphere in the cylinder and which locate above the xy plane is [tex]\mathbf{ a^2 ( \pi -2)}[/tex]
Step-by-step explanation:
The surface area of the sphere is:
[tex]\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA[/tex]
and the cylinder [tex]x^2 + y^2 =ax[/tex] can be written as:
[tex]r^2 = arcos \theta[/tex]
[tex]r = a cos \theta[/tex]
where;
D = domain of integration which spans between [tex]\{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}[/tex]
and;
the part of the sphere:
[tex]x^2 + y^2 + z^2 = a^2[/tex]
making z the subject of the formula, then :
[tex]z = \sqrt{a^2 - (x^2 +y^2)}[/tex]
Thus,
[tex]\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}[/tex]
[tex]\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}[/tex]
Similarly;
[tex]\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}[/tex]
[tex]\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}[/tex]
So;
[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}[/tex][tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}[/tex]
[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}[/tex]
[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}[/tex]
[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}[/tex]
From cylindrical coordinates; we have:
[tex]\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}[/tex]
dA = rdrdθ
By applying the symmetry in the x-axis, the area of the surface will be:
[tex]A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA[/tex]
[tex]A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta[/tex]
[tex]A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta[/tex]
[tex]A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta[/tex][tex]A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta[/tex]
[tex]A = 2a^2 [ cos \theta + \theta ]^{\dfrac{\pi}{2} }_{0}[/tex]
[tex]A = 2a^2 [ cos \dfrac{\pi}{2}+ \dfrac{\pi}{2} - cos (0)- (0)][/tex]
[tex]A = 2a^2 [0 + \dfrac{\pi}{2}-1+0][/tex]
[tex]A = a^2 \pi - 2a^2[/tex]
[tex]\mathbf{A = a^2 ( \pi -2)}[/tex]
Therefore, the area of the sphere in the cylinder and which locate above the xy plane is [tex]\mathbf{ a^2 ( \pi -2)}[/tex]
